Notes on compact Lie groups

Dietmar Salamon

ETH-Zurich

12 July 2013

Contents

1 Lie groups 1

2 Lie group homomorphisms 5

3 The Haar measure 7

4 Invariant inner products 11

5 Maximal toral subgroups 13

6 The center 17

7 Isotropy subgroups 22

8 Centralizers 23

9 Simple groups 25

10 Examples 27

1 Lie groups

A Lie Group is a smooth manifold G with a group structure such that themultiplication and the inverse map are smooth (C∞). The tangent space at

1

the identity element 1l ∈ G is called the Lie algebra of G and is denoted by

g := Lie(G) := T1lG.

For every g ∈ G the right and left multiplication maps Rg, Lg : G → G aredefined by

Rg(h) := hg, Lg(h) := gh

for h ∈ G. We shall denote the derivatives of these maps by

vg := dRg(h)v ∈ ThgG, gv := dLg(h)v ∈ TghG

for v ∈ ThG. In particular, for h = 1l and ξ ∈ T1lG = g, we have ξg, gξ ∈ TgGand hence ξ determines two vector fields g 7→ gξ and g 7→ ξg on G. Theseare called the left-invariant respectively right-invariant vector fields gen-erated by ξ. We shall prove in Lemma 1.2 below that the integral curves ofboth vector fields through g0 = 1l agree.

Exercise 1.1. (i) Prove that

(v0g1)g2 = v0(g1g2)

for v0 ∈ Tg0G and g1, g2 ∈ G. Similarly

(g0v1)g2 = g0(v1g2), (g0g1)v2 = g0(g1v2).

(ii) Prove that with the above notation the usual multiplication rule holds:If α, β : R → G are smooth curves then

d

dtα(t)β(t) = α(t)β(t) + α(t)β(t).

(Hint: Consider the partial derivatives of the map R2 → G : (s, t) 7→α(s)β(t) and use the chain rule.)

(iii) Deduce thatd

dtγ(t)−1 = −γ(t)−1γ(t)γ(t)−1

for every curve γ : R → G.

(iv) Prove that the vector fields g 7→ gξ and g 7→ ξg are complete for everyξ ∈ g. (Hint: Prove that the length of the existence interval is independentof the initial condition.)

2

Lemma 1.2. Let ξ ∈ g and γ : R → G be a smooth function. Then thefollowing conditions are equivalent.

(i) For all s, t ∈ R

γ(t+ s) = γ(s)γ(t), γ(0) = 1l, γ(0) = ξ. (1)

(ii) For all t ∈ R

γ(t) = ξγ(t), γ(0) = 1l. (2)

(iii) For all t ∈ R

γ(t) = γ(t)ξ, γ(0) = 1l. (3)

Moreover, for every ξ ∈ g there exists a unique smooth function γ : R → Gthat satisfies either of these conditions.

Proof. That (i) implies (ii) follows by differentiating the identity (1) with re-spect to s at s = 0. To prove that (ii) implies (i) note that, by Exercise 1.1 (i),the curves α(t) = γ(t + s) and β(t) = γ(t)γ(s) are both integral curves ofthe vector field g 7→ ξg such that α(0) = β(0) = γ(s). Hence they are equal.This shows that (i) is equivalent to (ii). That (i) is equivalent to (iii) followsby analogous arguments, interchanging s and t. The last assertion about theexistence of γ follows from Exercise 1.1 (iv).

The exponential map exp : g → G is defined by

exp(ξ) := γξ(1),

where γξ : R → G is the unique solution of (1). With this definition the pathγξ is given by

γξ(t) = exp(tξ).

To see this note that the curve α(s) = γξ(ts) satisfies α(s) = tξα(s). Hencethe exponential map satisfies

d

dtexp(tξ) = ξ exp(tξ) = exp(tξ)ξ.

The adjoint representation of G on its Lie algebra g is defined by

Ad(g)η := gηg−1 :=d

dt

∣∣∣∣t=0

g exp(tη)g−1.

3

In other words the linear map Ad(g) : g → g is the differential of the mapG → G : h 7→ ghg−1 at h = 1l. The map G → Aut(g) : g 7→ Ad(g) is a grouphomomorphism

Ad(gh) = Ad(g)Ad(h), Ad(1l) = id,

and is called the adjoint action of G on its Lie algebra. The differentialof this map at g = 1l in the direction ξ ∈ g is denoted by ad(ξ). The Liebracket of two elements ξ, η ∈ g is defined by

[ξ, η] := ad(ξ)η =d

dt

∣∣∣∣t=0

exp(tξ)η exp(−tξ).

Lemma 1.3. For all ξ, η, ζ ∈ g we have

[ξ, η] = −[η, ξ],

[[ξ, η], ζ ] + [[η, ζ ], ξ] + [[ζ, ξ], η] = 0.

Proof. We prove that the map g → Vect(G) : ξ 7→ Xξ defined by Xξ(g) := ξgis a Lie algebra homomorphism. To see this denote by ψt ∈ Diff(G) the flowgenerated by Xξ, that is ψt(g) := exp(tξ)g. Then, by definition of the Liebracket of vector fields,

[Xξ, Xη](g) =d

dt

∣∣∣∣t=0

dψt(ψ−t(g))Xη(ψ−t(g))

=d

dt

∣∣∣∣t=0

exp(tξ)η exp(−tξ)g

= [ξ, η]g

Here we have used Exercise 1.1 (i). Now the statement follows from theproperties of the Lie bracket for vector fields.

Lemma 1.4. Let ξ, η ∈ g and define γ : R → G by

γ(t) := exp(tξ) exp(tη) exp(−tξ) exp(−tη).

Then γ(0) = 0 and ddt

∣∣t=0

γ(√t) = [ξ, η].

Proof. As in the proof of Lemma 1.3, the flow of the vector field Xξ(g) = ξgon G is given by t 7→ Lexp(tξ) and [Xξ, Xη] = X[ξ,η] for ξ, η ∈ g. Hence theresult follows from the corresponding formula for general vector fields.

4

Lemma 1.5. Let R2 → G : (s, t) 7→ g(s, t) be a smooth function. Then

∂s(g−1∂tg

)− ∂t

(g−1∂sg

)+[g−1∂sg, g

−1∂tg]= 0.

Proof. If M is a smooth manifold, γ : R2 → M is a smooth function, andX(s, t), Y (s, t) ∈ Vect(M) are smooth families of vector fields such that

∂sγ = X γ, ∂tγ = Y γ,

then(∂sY − ∂tX − [X, Y ]) γ = 0.

To obtain the required formula, apply this to the manifold M = G, thefunction g : R2 → G, and the vector fields X(s, t) = Xξ(s,t), Y (s, t) = Xη(s,t),where ξ = (∂sg)g

−1 and η = (∂tg)g−1.

Exercise 1.6. Prove that for every g ∈ G and every ξ ∈ g

g exp(ξ)g−1 = exp(Ad(g)ξ)

(Hint: Consider the curve γ(t) = g exp(tξ)g−1 and use Exercise 1.1.)

Exercise 1.7. Prove that any two elements ξ, η ∈ g satisfy [ξ, η] = 0 if andonly if exp(sξ) and exp(tη) commute for all s, t ∈ R.

Exercise 1.8. Prove that Ad(exp(ξ)) = exp(ad(ξ)) for every ξ ∈ g. Hint:See Lemma 2.1 below.

2 Lie group homomorphisms

Let G and H be Lie groups with Lie algebras g and h. A Lie group homo-morphism is a smooth map φ : G → H which is a group homomorphism. Alinear map Φ : g → h is called a Lie algebra homomorphism if

[Φ(ξ),Φ(η)] = Φ([ξ, η])

for all ξ, η ∈ g. The next lemma asserts that the differential of a Lie grouphomomorphism at the identity is a Lie algebra homomorphism. An exampleis the map g → Vect(G) : ξ 7→ Xξ in the proof of Lemma 1.3; the corre-sponding Lie group homomorphism is the map G 7→ Diff(G) : g 7→ Lg. (SeeExample 10.14 below.)

5

Lemma 2.1. If φ : G → H is a Lie group homomorphism then Φ = dφ(1l) :g → h is a Lie algebra homomorphism

Proof. We show first that Φ and φ intertwine the the exponential maps, i.e.

exp(Φ(ξ)) = φ(exp(ξ)) (4)

for all ξ ∈ g. To see consider the curve γ(t) := φ(exp(tξ)) ∈ H. This curvesatisfies γ(s + t) = γ(s)γ(t) for all s, t ∈ R and γ(0) = Φ(ξ). Hence, byLemma 1.2, γ(t) = exp(tΦ(ξ)). With t = 1 this proves (4).

Next we prove that

Φ(g−1ξg) = φ(g)−1Φ(ξ)φ(g) (5)

for ξ ∈ g and g ∈ G. To see this, consider the curve γ(t) := g exp(tξ)g−1.By (4),

φ(γ(t)) = φ(g) exp(tΦ(ξ))φ(g)−1.

Differentiating this curve at t = 0 we obtain (5). It follows from (4) and (5)that

Φ([ξ, η]) =d

dt

∣∣∣∣t=0

Φ(exp(tξ)η exp(−tξ))

=d

dt

∣∣∣∣t=0

exp(tΦ(ξ))Φ(η) exp(−tΦ(ξ))

= [Φ(ξ),Φ(η)]

for all ξ, η ∈ g. This proves the lemma.

A representation of G is a Lie group homomorphism ρ : G → Aut(V )where V is a real or complex vector space. Differentiating such a map atg = 1l gives a Lie algebra homomorphism ρ : g → End(V ) defined by

ρ(ξ) :=d

dt

∣∣∣∣t=0

ρ(exp(tξ))

for ξ ∈ g. For example the map G → Aut(g) : g 7→ Ad(g) is the adjointrepresentation of G on its Lie algebra with corresponding Lie algebra ho-momorphism g 7→ End(g) : ξ 7→ ad(ξ). Also there is the obvious action ofU(n) on Cn and the induced actions on spaces of symmetric polynomials orexterior forms.

6

3 The Haar measure

Let G be a compact Lie group and denote by C(G) the space of continuousfunctions f : G → R with the norm

‖f‖ := supg∈G

|f(g)| .

The next theorem asserts the existence of a translation invariant measure onevery compact Lie group. The result and its proof extend to every compacttopological group that satisfies the second axiom of countability (i.e. it hasa finite or countable basis).

Theorem 3.1. Let G be a compact Lie group. Then there exists a boundedlinear functional M : C(G) → R that satisfies the following conditions.

(i) M(1) = 1.

(ii) M is left invariant, i.e. M(f Lg) =M(f) for f ∈ C(G) and g ∈ G.

(iii) M is right invariant, i.e. M(f Rg) =M(f) for f ∈ C(G) and g ∈ G.

(iv) If f ≥ 0 and f 6= 0 then M(f) > 0.

(v) Let φ : G → G denote the diffeomorphism defined by φ(g) = g−1. ThenM(f φ) =M(f) for every f ∈ C(G).

M is uniquely determined by (i) and either (ii) or (iii). It is called the Haarmeasure on G.

Proof. We follow notes by Moser which in turn are based on a proof byPontryagin. Let A denote the set of all measures on G of the form

A =

k∑

i=1

αiδai

where αi ∈ Q and∑

i αi = 1. If B =∑ℓ

j=1 βjδbj is another such measure,denote

A · B :=

k∑

i=1

ℓ∑

j=1

αiβjδaibj .

7

This defines a group structure onA. For A ∈ A we define two linear operatorsLA, RA : C(G) → C(G) by

(LAf)(g) :=m∑

i=1

αif(aig), (RAf)(g) :=m∑

i=1

αif(gai)

for f ∈ C(G) and g ∈ G. Then

LA(f Rh) = (LAf) Rh, RA(f Lh) = (RAf) Lh, (6)

LA·B = LB LA, RA·B = RA RB, LA RB = RB LA, (7)

min f ≤ LAf ≤ max f, min f ≤ RAf ≤ max f. (8)

We make use of the following three observations.

Observation 1: Denote Osc(f) := max f − min f . If f ∈ C(G) is non-constant then there exists an A ∈ A such that min f < minLAf and henceOsc(LAf) < Osc(f).

Suppose f assumes its maximum at a point g0 ∈ G. Choose a neighbourhoodU ⊂ G of 1l such that

gg0−1 ∈ U =⇒ f(g) >

1

2(max f +min f).

Now G =⋃

a∈G a−1U . Since G is compact, there exist finitely many points

a1, . . . , am ∈ G such that

G =

m⋃

i=1

ai−1U.

This means that for every h ∈ G there exists an i such that aih ∈ U .Consider the measure A := m−1

∑i δai . Since, for every g ∈ G, at least one

of the points aigg0−1 lies in U we obtain

(LAf)(g) =1

m

m∑

i=1

f(aig)

≥ m− 1

mmin f +

1

2m(max f +min f)

> min f.

Hence minLAf > min f and, by (8), Osc(LAf) < Osc(f).

8

Observation 2: For every f ∈ C(G) the set L(f) := LAf |A ∈ A isbounded and equicontinuous.

Boundedness follows from (8). To prove equicontinuity, note that, since Gis compact and second countable, it is a metrizable topological space. Letd : G × G → R be a distance function which induces the given topology.Fix a function f ∈ C(G) and an ε > 0. Since G is compact, f is uniformlycontinuous. Hence there is a δ > 0 such that, for all g, h ∈ G,

d(g, h) < δ =⇒ |f(g)− f(h)| < ε. (9)

We prove that there is an open neighbourhood U ⊂ G of 1l such that

g−1h ∈ U =⇒ d(g, h) < δ. (10)

We argue by contradiction. Suppose that there exist sequences gν , hν ∈ Gsuch that gν

−1hν → 1l and d(gν, hν) ≥ δ. Passing to a subsequence we mayassume that gν converges to g. Then hν = gν(gν

−1hν) converges also to g.Hence, for ν sufficiently large, we have d(gν, g) < δ/2 and d(hν, g) < δ/2,contradicting the assumption that d(gν , hν) ≥ δ. This proves (10). A similarargument shows that there is a constant δ′ > 0 such that

d(g, h) < δ′ =⇒ g−1h ∈ U. (11)

Now let g, h ∈ G such that d(g, h) < δ′. Then, by (11), g−1h ∈ U , hence(ag)−1(ah) = g−1h ∈ U for every a ∈ G, hence, by (10), d(ag, ah) < δ, andhence it follows from (9) that |f(ag) − f(ah)| < ε for every a ∈ G. Thisimplies |(LAf)(g) − (LAf)(h)| < ε for every A ∈ A. Thus we have provedequicontinuity.

Observation 3: For every f ∈ C(G), infA∈AOsc(LAf) = 0.

Choose a sequence Aν ∈ A such that

limν→∞

Osc(LAνf) = inf

A∈AOsc(LAf).

By Observation 2 and the Arzela-Ascoli theorem, the sequence fν := LAνf

has a uniformly convergent subsequence (still denoted by fν). Let f0 denotethe limit of this subsequence. Then

Osc(f0) = infA∈A

Osc(LAf). (12)

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Now, for every B ∈ A,

Osc(LBf0) = limν→∞

Osc(LBLAνf) = lim

ν→∞Osc(LAν ·Bf) ≥ Osc(f0).

The penultimate equality follows from (7) and the last inequality from (12).By Observation 1, f0 is constant. Hence Osc(f0) = 0 and so Observation 3follows from (12).

Observation 3 shows that there is a sequence Aν ∈ A such that LAνf

converges uniformly to a constant p ∈ R (called a left mean of f). Similarly,there exists a sequence Bν ∈ A such that RBν

f converges uniformly to aconstant q ∈ R (called a right mean of f). Since

‖LARBf − RBf‖ ≤ Osc(RBf), ‖RBLAf − LAf‖ ≤ Osc(LAf) (13)

for f ∈ C(G) and A,B ∈ A it follows that the right and left means agree andhence are independent of the choices of the sequences Aν and Bν . Namely,

p = limν→∞

RBνLAν

f = limν→∞

LAνRBν

f = q.

Let us define the operator M : C(G) → R by

M(f) := limν→∞

LAνf = lim

ν→∞RBν

f,

where Aν , Bν ∈ A are chosen such that Osc(LAνf) and Osc(RBν

f) convergeto zero. Thus M(f) is the left mean and the right mean of f . It is immediatefrom this definition that M(1) = 1, M(λf) = λM(f) for λ ∈ R, that M isleft and right invariant, and

min f ≤M(f) ≤ max f.

Now let f, f ′ ∈ C(M) and choose sequences Aν , Bν ∈ A such that

M(f) = limν→∞

LAνf, M(f ′) = lim

ν→∞RBν

f ′.

SinceM is left and right invariant, we haveM(LAνRBν

(f+f ′)) =M(f+f ′).Hence there is a sequence Cν ∈ A such that

M(f + f ′) = limν→∞

LCνRBν

LAν(f + f ′).

10

By (13), the right hand side also converges to M(f) +M(f ′) and hence

M(f + f ′) =M(f) +M(f ′)

for f, f ′ ∈ C(G). Thus we have proved thatM is a nonegative bounded linearfunctional that satisfies the assertions (i), (ii), and (iii) of the theorem.

We prove that M satisfies (iv). Hence let f ∈ C(G) be a function suchthat f ≥ 0 and f 6≡ 0. Then, by Observation 1, there exists an A ∈ A suchthat

minLAf > 0.

Choose Bν ∈ A such that RBνf converges to M(f). Then

M(f) = limν→∞

LARBνf = lim

ν→∞RBν

LAf ≥ minLAf > 0

as claimed.Next we prove that M is uniquely determined by conditions (i) and (ii).

To see this, let M ′ be another bounded linear functional on C(G) that satis-fies (i) and (ii). Then M ′(c) = c for every constant c and

M ′(LAf) =M ′(f)

for every f ∈ C(G) and every A ∈ A. Given f ∈ C(G) choose a sequenceAν ∈ A such that LAν

f converges uniformly to M(f). Then

M(f) =M ′(M(f)) = limν→∞

M ′(LAνf) =M ′(f).

This proves uniqueness. That M satisfies condition (v) follows from unique-ness and the fact that the map C(G) → R : f 7→ M(f φ) is a boundedlinear functional that satisfies (i) and (ii). This proves the theorem.

4 Invariant inner products

An inner product 〈., .〉 on g is called invariant if it is invariant under theadjoint action of G, i.e.

〈g−1ξg, g−1ηg〉 = 〈ξ, η〉for ξ, η ∈ g and g ∈ G. A Riemannian metric on G is called bi-invariantif the left and right translations Lh and Rh are isometries for every h ∈ G.Every invariant inner product on g determines a bi-invariant metric on G via

〈v, w〉 := 〈g−1v, g−1w〉 = 〈vg−1, wg−1〉 (14)

11

for v, w ∈ TgG. In turn, such a metric determines a volume form and hencea bi-invariant measure on G. By Theorem 3.1 this agrees with the Haarmeasure up to a constant factor. Conversely, if G is compact, one can usethe existence of a translation invariant measure to prove the existence of aninvariant inner product.

Proposition 4.1. Let G be a compact Lie group. Then g carries an invariantinner product.

Proof. LetM : C(G) → R denote the Haar measure and Q : g×g → R be anyinner product. For ξ, η ∈ g define fξ,η : G → R by fξ,η(g) := Q(gξg−1, gηg−1).Then the formula 〈ξ, η〉 :=M(fξ,η) defines an inner product on g. That it isinvariant follows from the formula fhξh−1,hηh−1 = fξ,η Rh.

Remark 4.2. (i) The proof of Proposition 4.1 shows that the existence of aright invariant measure on G suffices to establish the existence of an invariantinner product on g, and hence the existence of a bi-invariant measure on G.

(ii) On any Lie group the existence of a right invariant measure is easy toprove. Choose any inner product on g and extend it to a Riemannian metricon G by left translation. Then the right translations are isometries and hencethe volume form defines a right invariant measure on G.

(iii) Combining (i) and (ii) gives rise to a simpler proof of the existence of aHaar measure for compact Lie groups.

(iv) Uniqueness in Theorem 3.1 implies that every left invariant measure isright invariant. Here is a direct proof for compact Lie Groups: If ω is a leftinvariant volume form on G then so is Rg

∗ω. Hence there exists a grouphomomorphism λ : G → R such that Rg

∗ω = eλ(g)ω. Since G is compact, theonly group homomorphism from G to R is λ = 0.

Lemma 4.3. Let G be a compact Lie group with a bi-invariant Riemannianmetric. Then the geodesics have the form γ(t) = exp(tξ)g for g ∈ G andξ ∈ g.

Proof. Let I = [a, b] ⊂ R be a closed interval and γ0 : I → G be a geodesic.Let ξ : I → g be a smooth function such that ξ(a) = ξ(b) = 0 and considerthe function

γ(s, t) := γ0(t) exp(sξ(t)).

12

Then γ−1∂sγ = ξ and hence

1

2

d

ds

∫ b

a

〈γ−1∂tγ, γ−1∂tγ〉 dt =

∫ b

a

〈∂s(γ−1∂tγ), γ−1∂tγ〉 dt

=

∫ b

a

〈∂t(γ−1∂sγ), γ−1∂tγ〉 dt

= −∫ b

a

〈ξ, ∂t(γ−1∂tγ)〉 dt.

Here the penultimate equality follows from Lemma 1.5 and Exercise 4.5.Since the left hand side vanishes for s = 0 and every ξ it follows that∂t(γ0

−1∂tγ0) ≡ 0. This proves the lemma.

Exercise 4.4. (i) Prove that the group GL+(n,R) of real n × n-matriceswith positive determinant is connected.

(ii) Prove that the exponential map exp : Rn×n → GL+(n,R) is not surjec-tive. (Hint: Every negative eigenvalue of an exponential matrix Φ = exp(A)must have even multiplicity.)

(iii) Prove that Φ2 is an exponential matrix for every Φ ∈ GL(n,R).

(iv) Prove that for every compact connected Lie group G the exponentialmap exp : g → G is onto. (Hint: Use Proposition 4.1 (existence of aninvariant inner product), Lemma 4.3 (geodesics and exponential map), andthe Hopf-Rinow theorem (the existance of minimal geodesics).

Exercise 4.5. Let G be a compact connected Lie group. Prove that an innerproduct 〈·, ·〉 on g := Lie(G) is invariant if and only if

〈[ξ, η], ζ〉 = 〈ξ, [η, ζ ]〉for ξ, η, ζ ∈ g.

5 Maximal toral subgroups

Let G be a compact connected Lie group. A Lie subgroup of G is a closedsubgroup H which is a submanifold. A linear subspace h ⊂ g is called aLie subalgebra if it is invariant under the Lie bracket. If H ⊂ G is a Liesubgroup then, by definition of the Lie bracket, h := T1lH is a Lie subalgebraof g. A maximal torus in G is a connected abelian subgroup T ⊂ G whichis not properly contained in any other connected abelian subgroup. Thefundamental example is the subgroup of diagonal matrices in U(n) or SU(n).

13

Exercise 5.1. Let T ⊂ G be a maximal torus with Lie algebra t := Lie(T).Let η ∈ g such that [η, τ ] = 0 for every τ ∈ t. Prove that η ∈ t.

Lemma 5.2. Let G be a compact connected Lie group and T ⊂ G be amaximal torus. Then every element in G is conjugate to an element in T.

Proof. Given h ∈ G choose ξ ∈ g with exp(ξ) = h. Such an element exists byExercise 4.4 (iv). Then, by Exercise 1.6, ghg−1 = exp(gξg−1) for every g ∈ G.Hence we must find g ∈ G such that gξg−1 ∈ Lie(T) = t. Choose an invariantinner product on g and fix a generator τ ∈ t such that exp(sτ) | s ∈ R isdense in T . Since the orbit of ξ under the adjoint action of G is compactthere is an η ∈ g, conjugate to ξ, which minimizes the distance to τ in thisconjugacy class, i.e.

|η − τ |2 = infg∈G

|gηg−1 − τ |2.

We must prove that η ∈ t. To see this differentiate the map

G → R : g 7→ |gηg−1 − τ |2

at g = 1l to obtain 〈η − τ, [ζ, η]〉 = 0 for all ζ ∈ g. This implies 〈ζ, [η, τ ]〉 = 0for all ζ ∈ g and hence [η, τ ] = 0. By Exercise 1.7, exp(tη) commutes withexp(sτ) for all s and t. Since τ generates the torus, it follows that exp(tη)commutes with T for every t and hence [η, t] = 0. By Exercise 5.1, thisimplies η ∈ t.

Lemma 5.3. Any two maximal tori in G are conjugate.

Proof. Let T1,T2 ⊂ G be two maximal tori and choose an element g2 ∈ T2

such that T2 = cl(g2

k | k ∈ Z). By Lemma 5.2, there exists a g ∈ G such

that g2 ∈ gT1g−1. Hence T2 ⊂ gT1g

−1, and hence T2 = gT1g−1.

Lemma 5.3 shows that any two maximal tori in G have the same dimen-sion. This dimension is called the rank of G. The rank of G agrees with thedimension of a maximal abelian Lie subalgebra of g. (Prove this!)

Lemma 5.4. Let T ⊂ G be a maximal torus. Then every element of the Liealgebra g := Lie(G) is conjugate to an element of t := Lie(T).

Proof. Let ξ ∈ g. The the set exp(sξ) | s ∈ R is a torus and hence iscontained in a maximal torus T′. By Lemma 5.3, there exists a g ∈ Gsuch that gT′g−1 = T. Hence exp(sgξg−1) ∈ T for every s ∈ R, and hencegξg−1 ∈ t. This proves Lemma 5.4.

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Lemma 5.5. Let G be a compact connected Lie group and T ⊂ G be amaximal torus. Then T is a maximal abelian subgroup of G.

Proof. We follow the argument of Frank Adams in Lectures on Lie groups.Let h ∈ G be an element that commutes with T. We shall prove that h ∈ T.To see this let S ⊂ G be a maximal torus containing h and denote by

H := cl(hk | k ∈ Z

)

the subgroup of G generated by h. Examining closed subgroups of tori wesee that H is a Lie subgroup of S. Moreover, the Lie algebra h = Lie(H)commutes with t = Lie(T) and hence must be contained in t. Hence theidentity component of H is equal to H ∩ T and the quotient H/(H ∩ T) is afinite group. This finite group is generated by a single element [h] ∈ G/(T∩H)and hence is isomorphic to Zm for some integer m. This implies hm ∈ T.Hence the set

T := hit | t ∈ T, 1 ≤ i ≤ m− 1is a Lie subgroup of G such that

T/T ∼= Zm.

Any such group is generated by a single element h. To see this, let φ :Rn/Zn → T be an isomorphism, choose a vector ω = (ω1, . . . , ωn) ∈ Rn

such that the numbers 1, ω1, . . . , ωn are rationally independent. Choose τ =(τ1, . . . , τn) ∈ Rn such that φ(τ) = hm. Then the element

h := hφ((ω − τ)/m) ∈ G

generates T. By Lemma 5.2, there exists a maximal torus containing h andhence both h and T. Since T is a maximal torus it follows that h ∈ T. Thisproves Lemma 5.5.

Example 5.6. In general, a maximal abelian subgroup need not be a torus.For example the n× n-matrices with diagonal entries ±1 and determinant 1form a maximal abelian subgroup of G = SO(n).

For every maximal torus T ⊂ G denote

GT :=g ∈ G | g−1Tg = T

.

The quotient W := GT/T is called the Weyl group of T. The next lemmashows that every adjoint orbit in g intersects t/W in precisely one point.

15

Lemma 5.7. Let G be a compact connected Lie group and T ⊂ G be amaximal torus. Let ξ, η ∈ t. Then the following are equivalent.

(i) There exists a g ∈ G such that g−1ξg = η.

(ii) There exists a g ∈ GT such that g−1ξg = η.

Proof. That (ii) implies (i) is obvious. Hence suppose that g0−1ξg0 = η for

some g0 ∈ G. Choose sequences ξν , ην ∈ t such that ξν → ξ, ην → η, and

cl(exp(sξν) | s ∈ R) = cl(exp(tην) | t ∈ R) = T

for every ν. Choose gν ∈ G such that

∣∣gν−1ξνgν − ην∣∣ = inf

g∈G

∣∣g−1ξνg − ην∣∣ . (15)

Since |g0−1ξνg0 − ην | converges to zero it follows that

limν→∞

∣∣gν−1ξνgν − ην∣∣ = 0. (16)

Now differentiate the function g 7→ |g−1ξνg − ην |2 at g = gν . Then, by (15),we obtain

0 =1

2

d

dt

∣∣∣∣t=0

∣∣exp(−tζ)gν−1ξνgν exp(tζ)− ην∣∣2

= 〈gν−1ξνgν − ην , [gν−1ξνgν , ζ ]〉

= 〈[gν−1ξνgν, ην ], ζ〉

for every ζ ∈ g. Hence [gν−1ξνgν , ην ] = 0. Since ην generates the torus T

this implies gν−1ξνgν ∈ t. Since ξν generates the torus, this implies gν ∈ GT.

Passing to a convergent subsequence, we may assume that gν converges tosome element g ∈ GT. By (16), we have

g−1ξg − η = limν→∞

(gν

−1ξνgν − ην)= 0

and this proves Lemma 5.7.

16

6 The center

Let G be a connected Lie group. The subgroup

Z(G) := g ∈ G | gh = hg ∀h ∈ G

is called the center of G It is a Lie subgroup with corresponding Lie subal-gebra

Z(g) := ξ ∈ g | [ξ, η] = 0 ∀ η ∈ g .Note that Z(G) is a normal subgroup and the center of the quotient G/Z(G)is trivial. The following theorem is due to Herman Weyl.

Theorem 6.1. Let G be a compact connected Lie group. Then the first Bettinumber of G is given by dimH1(G;R) = dimZ(g).

Proof. The proof consists of three steps.

Step 1: Suppose G is equipped with a bi-invariant Riemannian metric. Then

∇vX(g) =

(dξ(g)v +

1

2[ξ(g), η]

)g,

where ξ : G → g, η ∈ g, v = ηg ∈ TgG, and X(g) = ξ(g)g.

Suppose first that ξ(g) ≡ ξ is constant. Then, by Lemma 4.3, the integralcurves of Xξ are geodesics. Hence

∇XξXξ = 0

for every ξ ∈ g. Replace ξ by ξ + η to obtain

∇XηXξ +∇Xξ

Xη = 0

for all ξ, η ∈ g. Since

∇XηXξ −∇Xξ

Xη = [Xξ, Xη] = X[ξ,η],

it follows that

∇XηXξ =

1

2X[ξ,η].

This proves Step 1 in the case where ξ : G → g is constant. The general caseis an immediate consequence.

17

Step 2: The Riemann curvature tensor of G is given by

R(ξg, ηg)ζg = −1

4[[ξ, η], ζ ]g

for g ∈ G and ξ, η, ζ ∈ g.

Consider the right invariant vector fields Xξ(g) = ξg for ξ ∈ g. By Step 1,

∇XηXξ =

1

2X[ξ,η].

Hence Step 2 follows by straight forward calculation from the identity

R(Xξ, Xη)Xζ = ∇Xξ∇Xη

Xζ −∇Xη∇Xξ

Xζ +∇[Xξ,Xη ]Xζ .

Step 3: We prove the theorem.

Let e1, . . . , ek be an orthonormal basis of g. Then, by Step 2, the Ricci tensorof G is given by

Ric(ξg, ηg) =k∑

i=1

〈R(eig, ξg)ηg, eig〉 =1

4

k∑

i=1

〈[ξ, ei], [η, ei]〉 (17)

Hence Ric(ξg, ξg) ≥ 0 with equality iff ξ ∈ Z(g). Now let α ∈ Ω1(G) andchoose ξ : G → g such that

αg(ηg) = 〈ξ(g), η〉.

The Bochner-Weitzenbock formula asserts that

‖dα‖2L2 + ‖d∗α‖2L2 = ‖∇α‖2L2 +

∫

G

Ric(α, α)dvol. (18)

Since Ric(α, α) ≥ 0, this shows that α is harmonic if and only if ∇α ≡ 0 andRic(α, α) ≡ 0. By (17) and Step 1 this means that

dξ(g)ηg =1

2[η, ξ(g)] = 0

for every g ∈ G and every η ∈ g. Equivalently, ξ : G → g is constantand takes values in the center of g. Thus we have proved that the space ofharmonic 1-forms can be identified with Z(g). This proves the theorem.

18

Theorem 6.2. Let G be a compact Lie group. Then the following holds.

(i) The fundamental group of G is abelian.

(ii) If Z(G) is finite then so is π1(G).

Proof. Assertion (i) holds for every topological group. To see this let α, β :[0, 1] → G be loops with

α(0) = α(1) = β(0) = β(1) = 1l.

Denote

α#β(t) :=

α(2t), if 0 ≤ t ≤ 1/2,

α(1)β(2t− 1), if 1/2 ≤ t ≤ 1.

(Here the term α(1) can be dropped, but the more general form will beneeded below.) Define

αs(t) :=

α(2t− s), if s/2 ≤ t ≤ (s+ 1)/2,

1l, otherwise,

and

βs(t) :=

β(2t+ s− 1), if (1− s)/2 ≤ t ≤ 1− s/2,

1l, otherwise,

for 0 ≤ s, t ≤ 1. Then γs(t) = αs(t)βs(t) is a homotopy from γ0 = α#β toγ1 = β#α. This proves (i). To prove (ii) note that, by (i), the fundamentalgroup

Γ := π1(G),

is abelian and, by Theorem 6.1,

Hom(Γ,R) ∼= H1(G;R) = 0.

This implies that Γ is finite. To see this, note first that Γ is finitely generated.Let γ1, . . . , γn ∈ Γ be generators. Since Γ is abelian, the set R ⊂ Zn of allinteger vectors m = (m1, . . . , mn) that satisfy

γ1m1 · · · γnmn = 1

form a subgroup of Zn and there is a natural isomorphism

Γ ∼= Zn/R.

Since Hom(Γ,R) = R⊥ = 0 it follows that R spans Rn. Hence the quotientRn/R is compact, and hence Γ ∼= Zn/R is a finite set.

19

Now let us denote by G the universal cover of G. In explicit terms,

G = γ : [0, 1] → G | γ(0) = 1l / ∼

where ∼ denotes homotopy with fixed endpoints. The projection

π : G → G

is given by π([γ]) = γ(1).

Proposition 6.3. Let G be a connected Lie group. Then

Z(G) = π−1(Z(G)).

Proof. Let α, β : [0, 1] → G such that α(0) = β(0) = 1l and α(1) ∈ Z(G).Define

αs(t) :=

α((1 + s)t), if 0 ≤ t ≤ 1/(s+ 1),

α(1), otherwise,

and

βs(t) :=

β((2t− s)/(2− s)), if s/2 ≤ t ≤ 1,

1l, otherwise,

for 0 ≤ s, t ≤ 1. Since α(1) ∈ Z(G), we have

α1β1 = β1α1 = α#β.

Moreover α0 = α and β0 = β. Hence both αβ and βα are homotopic to α#β.This proves that π−1(Z(G)) ⊂ Z(G). The converse inclusion is obvious.

The commutator subgroup [G,G] ⊂ G is defined as the smallest sub-group of G that contains all commutators [a, b] := aba−1b−1 for a, b ∈ G.Thus [G,G] is the subset of all products of finitely many such commutators.It is a normal subgroup of G.

Proposition 6.4. Let G be a compact connected Lie group. Then Z(G) isfinite if and only if [G,G] = G.

Proof. Consider the subbundle

E := (g, ξg) | ξ ⊥ Z(g) ⊂ TG.

20

By Exercise 4.5, the Lie bracket of any two right invariant vector fieldsXξ(g) = ξg andXη(g) = ηg is contained in E. Hence, by Frobenius’ theorem,E is integrable. Let H be the leaf of E through 1l, i.e.

H :=γ(1) | γ : [0, 1] → G, γ(0) = 1l, γ(t)−1γ(t) ⊥ Z(g)

.

If α, β : [0, 1] → G are paths that are tangent to E then so are αβ and α−1.Hence H is a subgroup of G. Next we prove that

[G,G] ⊂ H.

To see this note that, for every pair ξ, η ∈ g the curve

γ(t) := exp(tξ) exp(tη) exp(−tξ) exp(−tη)is tangent to E. Since the exponential map is surjective it follows thatevery commutator [a, b] = aba−1b−1 of two elements in G lies in H. Hence[G,G] ⊂ H. Next we prove that

H ⊂ [G,G]

To see this note that, by Exercise 4.5, the orthogonal complement of Z(g) isspanned by vectors of the form [ξ, η] for ξ, η ∈ g. Choose ξ1, . . . , ξk, η1, . . . , ηksuch that the vectors [ξi, ηi] form a basis of Z(g)⊥. For i = 1, . . . , k defineγi : R → [G,G] by

γi(t) := exp(√tξi) exp(

√tηi) exp(−

√tξi) exp(−

√tηi)

for t ≥ 0 and γi(t) := γi(−t) for t < 0. Then γi is continuously differentiableand γi(0) = [ξi, ηi]. Hence the function φ : Rk → [G,G], defined by

φ(t1, . . . , tk) := γ1(t1) · · ·γk(tk),is a continuously differentiable embedding near t = 0 and it is everywheretangent to E. Hence the image U0 of a sufficiently small neighbourhood of0 ∈ Rk under φ is a neighbourhood of 1l in H with respect to the intrinsictopology of H and it is contained in [G,G]. More generally, for every h ∈ Hthe set U = U0h ⊂ H is a neighbourhood of h with respect to the intrinsictopology and Uh−1 ⊂ [G,G]. Hence the sets H ∩ [G,G] and H \ [G,G] areboth open with respect to the intrinsic topology of H. Since H ∩ [G,G] 6= ∅it follows that H ⊂ [G,G], as claimed. Thus we have proved that [G,G] = His a leaf of the foliation determined by E. Hence [G,G] = G if and only ifE = TG if and only if Z(G) is finite. This proves Proposition 6.4.

21

Corollary 6.5. Let G be a compact connected Lie group with finite center.Then every principal G-bundle P → Σ over a compact oriented Riemannsurface of sufficiently large genus carries a flat connection.

Proof. By Theorem 6.2, π1(G) is finite, and hence G is compact. By Propo-sition 6.3, we have

π1(G) = π−1(1l) ⊂ Z(G).

There is a one-to-one correspondence between isomorphism classes of princi-pal G-bundles over a Riemann surface and elements γ ∈ π1(G). Suppose thatΣ is a Riemann surface of genus g and let Pγ → Σ be the principal bundlecorresponding to γ ∈ π1(G). Then a gauge equivalence class of flat connec-tion on Pγ (with respect to the identity component of the gauge group) canbe represented by elements

α1, . . . , αg, β1, . . . , βg ∈ G

that satisfyg∏

j=1

[αj , βj] = γ.

By Proposition 6.4, every element γ ∈ G can be expressed in this formwhenever Z(G) is finite. This proves the corollary.

7 Isotropy subgroups

Let G be a compact connected Lie group and M be a compact smoothmanifold equipped with a left action of G. The action will be denoted by

G×M → M : (g, x) 7→ gx.

The isotropy subgroup of an element x ∈ M is defined by

Gx := h ∈ G | hx = x .

Since Ggx = gGxg−1 the set of isotropy subgroups is invariant under conjuga-

tion. The next theorem asserts that the set of conjugacy classes of isotropysubgroups is finite.

Theorem 7.1. There exist finitely many Lie subgroups H1, . . . ,HN of G suchthat for every x ∈M there exists a j such that Gx is conjugate to Hj.

22

Proof. The proof is by induction on the dimension ofM . IfM is zero dimen-sional then the result is obvious. Now assume that dim M = n > 0 and thatthe result has been proved for all manifolds of dimensions less than n. Weprove that every point x0 ∈M has a neighbourhood U in which only finitelymany isotropy subgroups occur up to conjugacy. To see this, let G0 := Gx0

choose a G-invariant metric onM , denote by Lx : g → TxM the infinitesimalaction, and consider the horizontal space H0 := ker L∗

x0⊂ Tx0

M . Then theexponential map

G×H0 → M : (g, v0) 7→ g expx0(v0)

descends to a mapφ0 : G×G0

H0 →M,

where (g, v0) ∼ (gg0, g−10 v0) for g ∈ G, v0 ∈ H0, and g0 ∈ G0. The restriction

of φ0 to a sufficiently small neighbourhood of the zero section in the vectorbundle G ×G0

H0 → G/G0 is a G-equivariant diffeomorphism onto a neigh-bourhood of the G-orbit of x0. It follows that the isotropy groups of pointsx ∈M belonging to this neighbourhood are all conjugate to subgroups of G0

that appear as isotropy subgroups of the action of G0 on H0. By consider-ing the action of G0 on the unit sphere in H0 we obtain from the inductionhypothesis that there are only finitely many such isotropy subgroups. Thisproves the local statement. Cover M by finitely many such neighbourhoodsto prove the global assertion.

8 Centralizers

Let G be a compact connected Lie group. For any subset H ⊂ G the cen-tralizer of H is defined by

Z(H) := Z(H;G) := g ∈ G | gh = hg ∀h ∈ H

This set is a Lie subgroup of G with Lie algebra

Lie(Z(H)) =ξ ∈ g | hξh−1 = ξ ∀h ∈ H

=

⋂

h∈H

ker(1l− Ad(h)).

Moreover, Z(G) = Z(G;G) is the center of G and Z(Z(G);G) = G. Asubgroup H ⊂ G is abelian iff H ⊂ Z(H;G) and is maximal abelian iffH = Z(H ; G). For a singleton H = h we denote Z(h) := Z(h).

23

Lemma 8.1. Let G be a group. Then, for every subset H ⊂ G,

H ⊂ Z(Z(H)), Z(Z(Z(H)) = Z(H).

Proof. The first assertion follows directly from the definition and the secondfollows from the first. Namely, since H ⊂ Z(Z(H)) we have Z(Z(Z(H)) ⊂Z(H), and the converse inclusion follows by applying the first assertion toZ(H) instead of H. This proves Lemma 8.1.

A subgroup H ⊂ G is called a centralizer subgroup if there exists asubset of G whose centralizer is equal to H. By Lemma 8.1 this condition isequivalent to

H = Z(Z(H)). (19)

A Lie subalgebra h ⊂ g is called a centralizer subalgebra if there exists acentralizer subgroup H ⊂ G such that h = Lie(H). Let us denote by Z ⊂ 2G

the set of all centralizer subgroups. By (19), the map

Z → Z : H 7→ Z(H)

is an involution. Moreover the group G acts on Z by conjugation and theinvolution H 7→ Z(H) is equivariant under this action, i.e.

Z(gHg−1) = gZ(H)g−1.

The fixed points of the involution are the maximal abelian subgroups of G andhence are also fixed points of the conjugate action. Consider the equivalencerelation H ∼ H′ iff H′ = gHg−1 for some g ∈ G. The following theoremasserts that the quotient Z/ ∼ is a finite set.

Theorem 8.2. Let G be a compact connected Lie group. Then there existfinitely many centralizer subgroups H1, . . . ,Hm of G such that every central-izer subgroup H ⊂ G is conjugate to one of the Hi.

Proof. Since G is compact it admits a faithful representation ρ : G → U(n).Now let H ⊂ G be a subgroup and g ∈ Z(H). Then ρ(g) commutes with allmatrices in the span of ρ(H). Thus it suffices to pick n2 elements h1, . . . , hn2 ∈H such that ρ(H) is contained in the span of the matrices ρ(hi). Then Z(H)can be characterized as the set of all g ∈ G such that ρ(g) commutes withρ(hi) for i = 1, . . . , n2. In other words, if the group G acts on the vectorspace V := (Cn×n)n

2

by g · Ai := ρ(g)Aiρ(g)−1 for i = 1, . . . , n2, then every

centralizer subgroup of G is the isotropy subgroup of some element of V . ByTheorem 7.1, the set of conjugacy classes of such isotropy subgroups is finite.This proves Theorem 8.2.

24

9 Simple groups

A Lie subalgebra h ⊂ g is called an ideal if [h, g] ⊂ h. The Lie algebra of anormal Lie subgroup of G is necessarily an ideal. A Lie algebra g is calledsimple if it has no nontrivial ideals (that is 0 and g are the only ideals ing). It is called semi-simple if it is a direct sum of simple Lie algebras. ALie group is called simple (respectively semi-simple) if its Lie algebra issimple (respectively semi-simple).

Theorem 9.1. Every compact connected simply connected simple Lie groupis isomorphic to one in the following list

An := SU(n + 1), n ≥ 1,Bn := Spin(2n+ 1), n ≥ 2,Cn := Sp(n), n ≥ 3,Dn := Spin(2n), n ≥ 4,

or to one of the exceptional groups G2, F4, E6, E7, E8.

There are relations such as Spin(3) ∼= SU(2) ∼= Sp(1), Spin(5) ∼= Sp(2),and Spin(6) ∼= SU(4). The Lie groups D1

∼= Spin(2) ∼= U(1) ∼= S1 andSpin(4) ∼= SU(2) × SU(2) are not simple. (See Exercises 10.10 and 10.11below for Spin(3) and Spin(4).)

The Killing form

Every Lie algebra carries a natural pairing

κ(ξ, η) := trace(ad(ξ)ad(η))

called the Killing form. On su(n) this form is negative definite. In generalthe Killing form may have a kernel and/or be indefinite.

Theorem 9.2 (Cartan). The Killing form is nondegenerate (and negativedefinite) if and only if G is semisimple.

Exercise 9.3. Prove that the Killing forms on su(n) and so(2n) are givenby

κ(ξ, η) = −(2n− 1) trace(ξ∗η), ξ, η ∈ su(n),

κ(ξ, η) = −(n− 2) trace(ξTη), ξ, η ∈ so(2n).

Exercise 9.4. Prove that the Killing form on SL(2,R) is indefinite.

25

Root systems

Let G be a compact Lie group with maximal torus T . The exponential mapis onto by Exercise 4.4 (iv). It determines an isomorphism

T ∼= t/Λ

where t = Lie(T ) and

Λ := τ ∈ t | exp(τ) = 1l

is a lattice which spans t. A one-dimensional complex representation is ahomomorphism T → S1. Under the identification T ∼= t/Λ any such ho-momorphism is of the form τ 7→ e2πiw(τ) where w : t → R is a linear mapwith

w(Λ) ⊂ Z.

Any such map w ∈ t∗ is called a weight. Now consider the adjoint represen-tation of T on g. Since the action preserves any invariant inner product thecommuting endomorphisms Ad(τ) for τ ∈ t are simultaneously diagonalizable(over C). It follows that there exists a decomposition

g = t⊕⊕

α

Vα

where Vα ⊂ g are two dimensional representations of T . In other words thereexists a complex structure Jα on Vα and weights wα ∈ t∗ such that

[τ, ξ] = 2πJαwα(τ)ξ, τ ∈ t, ξ ∈ Vα.

The weights wα are called the roots of the Lie algebra g. For each α defineτα ∈ t to be the dual element with respect to the Killing form:

κ(τα, σ) = wα(σ), σ ∈ t.

The length of the root wα is defined by

ℓ(α) :=√−κ(τα, τα).

The length of the longest root is an important invariant of the Lie group G.We denote the square of its inverse by

a(G) :=1

supα ℓ(α)2.

26

Here is a list of these invariants for the simple groups.

G dim(G) a(G)

SU(n) n2 − 1 nSpin(n) 1

2n(n− 1) n− 2

Sp(n) n(2n + 1) n+ 1G2 14 4F4 52 9E6 78 12E7 133 18E8 248 30

10 Examples

Example 10.1 (General linear group). The group GL(n,R) of invertiblereal n × n-matrices is a Lie group. This space is an open set in Rn×n andhence is obviously a manifold. Its Lie algebra is the vector space Rn×n of allreal n× n matrices with Lie bracket operation

[A,B] := AB −BA.

In this case the exponential map

exp : Rn×n → GL(n,R)

is the usual exponential map for matrices and the expressions gv and vgfor v ∈ ThG are given by matrix multiplication. The example GL(n,C) ofinvertible complex n × n-matrices is similar. However, the group GL(n,C)is connected while the group GL(n,R) has two components distinguished bythe sign of the determinant.

Example 10.2 (Special linear group). The determinant map

det : GL(n,C) → C∗

is a Lie group homomorphism and its kernel is a Lie group denoted by

SL(n,C) :=Φ ∈ Cn×n | det Φ = 1

.

27

The formula det(exp(A)) = exp(trace(A)) shows that the Lie algebra ofSL(n,C) is given by

sl(n,C) :=A ∈ Cn×n | traceA = 0

.

The Lie group SL(n,R) with Lie algebra sl(n,R) is defined analogously.

Example 10.3 (Circle). The unit circle S1 := z ∈ C | |z| = 1 in thecomplex plane is a Lie group (under multiplication of complex numbers). ItsLie algebra is the space iR of imaginary numbers with zero Lie bracket. (SeeExercise 1.7.) There is a Lie group isomorphism

R/Z → S1 : t 7→ e2πit.

Example 10.4 (Torus). Let V be an n-dimensional real vector space andΛ ⊂ V be a lattice (a discrete additive subgroup) which spans V . Then

T := V/Λ

is a compact abelian Lie group (the group operation is the addition in V ) withLie algebra V . The exponential map is the projection V → V/Λ. Any suchLie group is called a torus. Tori can be characterized as compact connectedfinite dimensional abelian Lie groups. The basic example is the standardtorus

Tn := Rn/Zn

and every n-dimensional torus is isomorphic to Tn.

Example 10.5 (Orthogonal group). The orthogonal n×n-matrices forma Lie group

O(n) :=Φ ∈ Rn×n |ΦTΦ = 1l

.

This group has two components distinguished by the determinant det Φ = ±1and the component of the identity is denoted by

SO(n) := Φ ∈ O(n) | det Φ = 1.

Its Lie algebra is the space of antisymmetric matrices

so(n) :=A ∈ Rn×n |AT + A = 0

.

The group SO(n) is compact and connected and the exponential map issurjective (see Exercise 4.4).

28

Example 10.6 (Unitary group). The unitary n × n-matrices form a Liegroup

U(n) :=U ∈ Cn×n |U∗U = 1l

where U∗ denotes the conjugate transpose of U . This group is connected andits Lie algebra is given by

u(n) :=A ∈ Cn×n |A∗ + A = 0

.

The case n = 1 corresponds to the circle S1 = U(1). The subgroup of unitarymatrices of determinant 1 is denoted by

SU(n) := U(n) ∩ SL(n,C)

and its Lie algebra by

su(n) := A ∈ u(n) | trace(A) = 0 .

Both groups U(n) and SU(n) are compact and connected.

Example 10.7 (Unit quaternions). Denote by H = R4 the space ofquaternions

x = x0 + ix1 + jx2 + kx3

with (noncommutative) multiplicative structure

i2 = j2 = k2 = −1, ij = −ji = k, jk = −kj = i ki = −ik = j.

The norm of x ∈ H is defined by

|x|2 := xx = x02 + x1

2 + x22 + x3

2, x := x0 − ix1 − jx2 − kx3,

and satisfies the rule |xy| = |x| · |y|. Hence the unit quaternions form a group

Sp(1) := x ∈ H | |x| = 1

with unit 1 and inverse map x 7→ x. Its Lie algebra consists of the imaginaryquaternions

sp(1) := x ∈ H | x0 = 0 .The exponential map is given by the usual formula exp(x) =

∑∞

k=0 xk/k!.

The quaternion multiplication defines a group structure on S3 = Sp(1) and aLie algebra structure on R3 ≃ sp(1). This Lie algebra structure correspondsto the vector product.

29

Example 10.8. The quaternion matrices Φ ∈ Hn×n with Φ∗Φ = 1l form acompact connected group denoted by Sp(n). Its Lie algebra sp(n) consistsof the quaternion matrices A ∈ Hn×n with A∗ +A = 0. Here A∗ denotes theconjugate transpose as in the complex case.

Exercise 10.9. (i) Prove that the map Sp(1) → SU(2) : x 7→ U(x) definedby

U(x) :=

(x0 + ix1 x2 + ix3

−x2 + ix3 x0 − ix1

)

is a Lie group isomorphism.

(ii) Prove that the corresponding Lie algebra homomorphism sp(1) → su(2) :ξ 7→ u(ξ) is given by

u(ξ) :=

(iξ1 ξ2 + iξ3

−ξ2 + iξ3 −iξ1

).

Show that the matrices

I =

(i 00 −i

), J =

(0 1

−1 0

), K =

(0 ii 0

).

satisfy the quaternion relations. In other words, the Lie algebra su(2) isisomorphic to the imaginary quaternions and the isomorphism is given byi 7→ I, j 7→ J , k 7→ K. The natural orientation of SU(2) is determined bythe basis I, J,K of su(2).

(iii) Prove that

[u(ξ), u(η)] = 2u(ξ × η), trace(u(ξ)∗u(η)) = 2〈ξ, η〉

for ξ, η ∈ R3 ∼= Im(H).

Exercise 10.10 (Spin(3)). The unit quaternions act on the imaginary qua-ternions by conjugation. This determines a homomorphism Sp(1) → SO(3) :x 7→ Φ(x) defined by

Φ(x)ξ := xξx

for x ∈ Sp(1) and ξ ∈ Im(H) ∼= R3. On the left the multiplication isunderstood as a product of matrix and vector and on the right as a productof quaternions.

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(i) Prove that

Φ(x) =

x20 + x21 − x22 − x23 2(x1x2 − x0x3) 2(x0x2 − x1x3)2(x0x3 − x1x2) x20 − x21 + x22 − x23 2(x2x3 − x0x1)2(x1x3 − x0x2) 2(x0x1 − x2x3) x20 − x21 − x22 + x23

.

(ii) Verify that the map SU(2) → SO(3) : U(x) 7→ Φ(x) is a Lie grouphomomorphism and a double cover. Deduce that π1(SO(3)) = Z2.

(iii) Let su(2) → so(3) : u(ξ) 7→ A(ξ) denote the corresponding Lie algebrahomomorphism. Prove that

A(ξ) = 2

0 −ξ3 ξ2ξ3 0 −ξ1

−ξ2 ξ1 0

.

Prove that [A(ξ), A(η)] = 2A(ξ × η) and trace(A(ξ)TA(η)) = 8〈ξ, η〉.Exercise 10.11 (Spin(4)). The group Sp(1) × Sp(1) acts on H by the or-thogonal transformations x 7→ uxv for (u, v) ∈ Sp(1) × Sp(1). Prove thatthis axtion determines a double cover Sp(1) × Sp(1) → SO(4) and find anexplicit formula for the matrix Ψ(u, v) ∈ R4×4 defined by Ψ(u, v)x = uxv.

Lemma 10.12. (i) SO(n) is connected and in the case n ≥ 3 its fundamentalgroup is isomorphic to Z2. Hence for n ≥ 3 the universal cover of SO(n) isa compact group (with the same Lie algebra). It is denoted by Spin(n).

(ii) SU(n) is connected and simply connected and π2(SU(n)) = 0.

(iii) The fundamental group of U(n) is isomorphic to the integers. Thedeterminant homomorphism det : U(n) → S1 induces an isomorphism offundamental groups.

Proof. The subgroup of all matrices Φ ∈ SO(n) whose first column is thefirst unit vector e1 = (1, 0, . . . , 0) ∈ Rn is isomorphic to SO(n − 1). Hencethere is a fibration SO(n−1) → SO(n) → Sn−1 where the second map sendsa matrix in SO(n) to its first column. The homotopy exact sequence of thisfibration has the form

πk+1(Sn−1) → πk(SO(n− 1)) → πk(SO(n)) → πk(S

n−1)

By Exercise 10.10, π1(SO(3)) ≃ Z2. For n ≥ 4 this follows from the exactsequence with k = 1. The connectedness of SO(n) is obvious for n = 1, 2.For n ≥ 3 it follows from the exact sequence with k = 0. This proves (i).

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To prove (ii) consider the fibration SU(n − 1) → SU(n) → S2n−1 wherethe last map sends U ∈ SU(n) to the first column of U . The homotopy exactsequence of this fibration has the form

πk+1(S2n−1) → πk(SU(n− 1)) → πk(SU(n)) → πk(S

2n−1).

For n = 1 the group SU(1) = 1 is obviously connected and simply con-nected. For n ≥ 2 use the exact sequence inductively (over n) with k = 0, 1.The statement about π2 is proved similarly with k = 2.

To prove (iii) consider the fibration SU(n) → U(n) → S1. The homotopyexact sequence of this fibration has the form

1 = π1(SU(n)) → π1(U(n)) → π1(S1) → π0(SU(n)) = 1.

In view of statement (ii) this shows that π1(U(n)) ≃ π1(S1) ≃ Z.

Let Y be a compact oriented smooth 3-manifold, and recall from Exer-cise 10.9 that SU(2) is diffeomorphic to S3 and carries a natural orientation.Hence every smooth map g : Y → SU(2) has a well defined degree. Thenext proposition shows that this degree can be expressed as as the integralof natural 3-form over Y .

Lemma 10.13. For every compact oriented smooth 3-manifold Y and everysmooth map g : Y → SU(2) we have

∫

Y

trace(g−1dg ∧ g−1dg ∧ g−1dg

)= −24π2 deg(g).

Proof. Denote

ωg := trace(g−1dg ∧ g−1dg ∧ g−1dg

)∈ Ω3(Y ).

If f : Y ′ → Y is a smooth map then ωgf = f ∗ωg. In particular, withω0 := ωid ∈ Ω3(SU(2)), we have ωg = g∗ω0 and hence

∫

Y

ωg = deg(g)

∫

SU(2)

ω0. (20)

To compute the integral of ω0 consider the diffeomorphism U : S3 → SU(2)defined in Exercise 10.9. With the standard orientations of S3 and SU(2)this map has degree 1. Moreover, it follows from the symmetry of this map

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that ωU is a constant multiple of the volume form on S3. To find out thefactor we compute the form on the tangent space TxS

3 for x = (1, 0, 0, 0).On this space

U−1dU = Idx1 + Jdx2 +Kdx3,

hence

U−1dU ∧ U−1dU = 2Idx2 ∧ dx3 + 2Jdx3 ∧ dx1 + 2Kdx1 ∧ dx2,and hence

U−1dU ∧ U−1dU ∧ U−1dU = 2(I2 + J2 +K2)dx1 ∧ dx2 ∧ dx3This implies ωU = −12 dvolS3 and hence, by (20) with g = U ,

∫

SU(2)

ω0 =

∫

S3

ωU = −12Vol(S3) = −24π2.

This proves the proposition.

Example 10.14 (Diffeomorphisms). LetM be a compact manifold. Thenthe diffeomorphisms of M form an infinite dimensional Lie group Diff(M)with group multiplication given by composition (f, g) 7→ f g. Its Lie algebrais the space Vect(M) of vector fields on M . The Lie algebra structure onVect(M) is the usual one if the sign in the definition of the Lie bracket of twovector fields is chosen appropriately. The one-parameter subgroup generatedby a vector field X ∈ Vect(M) is its flow R → Diff(M) : t 7→ φt defined by

d

dtφt = X φt, φ0 = id.

Note also that the inverse of the adjoint action of Diff(M) on Vect(M) isgiven by pullback, i.e.

Ad(φ−1)X = φ∗X = dφ X φ−1.

Interesting subgroups are given by the volume preserving diffeomorphisms orthe isometries on a Riemannian manifold, or by the symplectomorphisms ona symplectic manifold.

Example 10.15 (Invertible linear operators). For invertible operatorson an infinite dimensional Hilbert space H the relation between Lie-groupand Lie-algebra is somewhat subtle. Not every one parameter group t 7→ S(t)of invertible linear operators is differentiable. Such groups can be generatedby unbounded operators and this leads to the theory of semigroups of linearoperators.

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